# Solubility of Metal Hydroxides

Learning Objectives

• Estimate the pH of the solution due to precipitate of a metal hydroxide.
• Calculate the maximum metal ion concentration when the pH is known.
• Explain behavior of amphoteric metal hydroxides.

Most metal hydroxides are insoluble; some such as $$\ce{Ca(OH)2}$$, $$\ce{Mg(OH)2}$$, $$\ce{Fe(OH)2}$$, $$\ce{Al(OH)3}$$ etc. are sparingly soluble. However, alkali metal hydroxides $$\ce{CsOH}$$, $$\ce{KOH}$$, and $$\ce{NaOH}$$ are very soluble, making them strong bases. When dissolved, these hydroxides are completely ionized. Since the hydroxide concentration, $$\ce{[OH- ]}$$, is an integrated property of the solution, the solubility of metal hydroxide depends on pH, pOH or $$\ce{[OH- ]}$$.

Alkali metal hydroxides $$\ce{LiOH}$$, $$\ce{NaOH}$$, $$\ce{KOH}$$, $$\ce{CsOH}$$ are soluble, and their solutions are basic. Hydroxides of alkali earth metals are much less soluble. For example, quicklime ($$\ce{CaO}$$) reacts with water to give slaked lime, which is slightly soluble.

$\begin{array}{ccccl} \ce{CaO &+ &H2O &\rightleftharpoons &\:Ca(OH)2}\\ \textrm{quicklime} &&&&\textrm{slaked lime (slightly soluble)} \end{array}$

Milk of magnesia is $$\ce{Mg(OH)2}$$ (Ksp = 7e-12) suspension. In an acidic solution such as stomach juice, the following reaction takes place,

$\ce{Mg(OH)2 + H+ \rightleftharpoons Mg^2+ + 2 H2O}$

Thus, it can neutralize excess acid in the stomach.

Example 1

Calculate the maximum concentration of $$\ce{Mg^2+}$$ in a solution which contains a buffer so that pH = 3 at 298 K.

Solution

As usual, we write the equilibrium equation so that we can write the concentration below the formula. If we do not know the concentration, we assume it to be a variable x.

$\begin{array}{ccccc} \ce{Mg(OH)2 &\rightleftharpoons &Mg^2+ &+ &2 OH-}\\ &&x && 1 \times 10^{-11} \end{array}$

$K_{\ce{sp}} = x (1 \times 10^{-11})^2 = 7 \times 10^{-12}$

Solving for x results in $$x = 7 \times ^{10}$$

DISCUSSION

This value certainly is too large, unrealistic.

Example 2

Calculate the pH of a saturated $$\ce{Mg(OH)2}$$ solution.

Solution

We assume the concentration to be x M of $$\ce{Mg(OH)}$$, and note that $$\ce{[OH- ]} = 2 x$$,

$\begin{array}{ccccc} \ce{Mg(OH)2 &\rightleftharpoons &Mg^2+ &+ &2 OH-}\\ &&x &&2 x \end{array}$

$K_{\ce{sp}} = x (2 x)^2 = \textrm{7e-12}$

Solving for x; x = 1.2e-4

\begin{align} \ce{[OH]} &= \textrm{2.4e-4}\\ \ce{pOH} &= 3.62 \end{align}

$\mathrm{pH = 14 - 3.62 = 10.38}$

DISCUSSION

The pH of a saturated lime ($$\ce{Ca(OH)2}$$) solution is about 10.0.

### Amphoteric Hydroxides

Not all metal hydroxides behave the same way - that is precipitate as hydroxide solids. Metal hydroxides such as $$\ce{Fe(OH)3}$$ and $$\ce{Al(OH)3}$$ react with acids and bases, and they are called amphoteric hydroxide. In reality, $$\ce{Al(OH)3}$$ should be formulated as $$\ce{Al(H2O)3(OH)3}$$, and this neutral substance has a very low solubility. It reacts in the following way as $$\ce{[H+]}$$ increases.

\begin{align} \ce{Al(H2O)3(OH)3 + H3O+ &\rightleftharpoons Al(H2O)4(OH)2+ + HOH}\\ \ce{Al(H2O)4(OH)2+ + H3O+ &\rightleftharpoons Al(H2O)5(OH)^2+ + H2O}\\ \ce{Al(H2O)5(OH)^2+ + H3O+ &\rightleftharpoons Al(H2O)6^3+ + H2O} \end{align}

When the pH increases, the following reactions take place:

\begin{align} \ce{Al(H2O)3(OH)3 + OH- &\rightleftharpoons Al(H2O)2(OH)4- + H2O}\\ \ce{Al(H2O)2(OH)4- + OH- &\rightleftharpoons Al(H2O)(OH)5^2- + H2O}\\ \ce{Al(H2O)(OH)5^2- + OH- &\rightleftharpoons Al(OH)6^3- + H2O} \end{align}

The charged species are soluble in water. As a result, amphoteric hydroxides dissolve in acidic and basic solutions.

## Questions

1. Assume the pH of gastric juice to be 2. Calculate the maximum $$\ce{[Mg^2+]}$$.
2. Calculate the pH of a 0.10 M $$\ce{NH3}$$ solution.
3. Calculate the maximum $$\ce{[Fe^2+]}$$ in a 0.10 M $$\ce{NH3}$$ solution. Give the value in M .
4. What are amphoteric metal hydroxides? (enter no more than one line)

## Solutions

1. Answer $$\mathrm{[Mg^{2+}] = 7e12\: M}$$
Consider...
We assume the temperature to be 298 K, which is too low.

$$\ce{[Mg^2+]} = \dfrac{\textrm{7e-12}}{(\textrm{1e-12})^2} =\: ?$$

This value is unrealistically large. The result is correct, but meaningless.

Consider...
$$\ce{[OH- ]} = (0.10\times\textrm{1.8e-5})^{1/2} = \textrm{1.34e-3}$$
$$\ce{[H+]} = \dfrac{\textrm{1e-14}}{\textrm{1.34e-3}} = \textrm{7.5e-12}$$, $$\ce{pH} =\: ?$$
This value is required for the calculation in next question. Better yet, remember that $$\ce{[OH- ]} = \textrm{1.34e-4}$$.
Consider...
$$\ce{[Fe^2+]} = \dfrac{\textrm{7.9e-16}}{(\textrm{1.34e-3})^2} =\: ?\:\ce M$$
What is the value in g/L? Molar mass of $$\ce{Fe}$$ is 55.8 g/mol.
4. Answer Metal hydroxides such as $$\ce{Fe(OH)3}$$ and $$\ce{Al(OH)3}$$ that react with acids and bases are called amphoteric hydroxide.